# Asymptotic Analysis question.

1. Jan 10, 2009

### MathematicalPhysicist

I have this question (it's not a coursework question, I'm learning it by my own):
Use an appropiate power series expansion to find an asymptotic approximation as $$\epsilon$$ approaches 0+, correct to O(epsilon^2), for the two small roots of the equation: $$\epsilon x^3+x^2+2x-3=0$$ Then by using a suitable rescaling, find the first three terms of an asymptotic expansion as epsilon approaches 0+ of the singular root.

My way to go around this is to write down the next expansion series (because of the degree of the equation):
$$x(\epsilon)=\frac{1}{\epsilon^3}x_{-3}+ \frac{1}{\epsilon^2}x_{-2}+ \frac{1}{\epsilon^1}x_{-1}+x_0+x1\epsilon+...$$, the singular root is when epsilon equals null, i.e to solve x^2+2x-3=0=(x+3)(x-1) the roots are: x=-3 and x=1, I guess I need to choose the second root, now the recursive equation is:
$$x^2_{n+1}=-x_{n}/\epsilon-2/\epsilon+3/(x_{n}\epsilon)$$ Now, x0=1 and then use the iterative equation in order to find x_-3 and x_-2, am I correct or totally way off here?

2. Jan 11, 2009

### MathematicalPhysicist

Anyone?

3. Jan 11, 2009

### lurflurf

h*x^3+x^2+2x-3=0
h=-1/x-2/x^2+3/x^3
h=-1/x
x=-1/h
x=-1/h+2+ax+bx^2+...
find more terms

4. Jan 11, 2009

### tim_lou

I don't quite understand what your method is... but if I were to tackle this problem, i'll get rid of the messy ϵ in x^3, and write:
$$g(x)=f\left(\frac{1}{x}\right)=\frac{1}{x^3}(-3x^3+2x^2+x+\epsilon)$$
(the motivation is that as ϵ goes to 0, the extra root goes to infinity. This is not good. So plugging in 1/x makes the root go to zero instead).

so, instead of finding roots of f, I find roots of g (much easier)
g=0 implies
$$x(3x^2-2x-1)=\epsilon$$
i'll just expand around -1/3, 1/2, 0 and solve this eq order by order.

at the end, just take the inverse and expand using geometric series.